Continuity

1. Definition of Continuity at a Point

A function $f(x)$ is said to be continuous at a point $x=a$ in its domain if the following three conditions are satisfied:

1. $f(a)$ is defined: The function has a finite value at $x=a$.
2. $\lim_{x \to a} f(x)$ exists: The limit of the function as $x$ approaches $a$ exists. This implies that the Left-Hand Limit (LHL) and Right-Hand Limit (RHL) are equal. $$\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$$
3. $\lim_{x \to a} f(x) = f(a)$: The limit of the function at the point is equal to the value of the function at that point.

Geometrically, a function is continuous at a point if its graph has no breaks, jumps, or holes at that point. You can draw the graph through the point without lifting your pen.

2. Continuity in an Interval

• Open Interval: A function is continuous in an open interval $(a, b)$ if it is continuous at every point in the interval.
• Closed Interval: A function $f(x)$ is continuous in a closed interval $[a, b]$ if:
  1. It is continuous in the open interval $(a, b)$.
  2. It is continuous from the right at $a$, i.e., $\lim_{x \to a^+} f(x) = f(a)$.
  3. It is continuous from the left at $b$, i.e., $\lim_{x \to b^-} f(x) = f(b)$.

3. Types of Discontinuities

If a function is not continuous at a point $x=a$, it is said to be discontinuous at that point. There are several types:

1. Removable Discontinuity: The limit $\lim_{x \to a} f(x)$ exists but is not equal to $f(a)$, or $f(a)$ is not defined.

   • Missing Point: $f(a)$ is not defined.
   • Isolated Point: $\lim_{x \to a} f(x) \neq f(a)$. This type of discontinuity can be 'removed' by redefining the function at the single point $x=a$.

2. Non-Removable Discontinuity (or Discontinuity of the First Kind): The limit $\lim_{x \to a} f(x)$ does not exist because the LHL and RHL exist and are finite, but they are not equal ($ext{LHL} \neq ext{RHL}$). This is also called a Jump Discontinuity.

3. Discontinuity of the Second Kind: Either the LHL or the RHL (or both) do not exist or are infinite. This is also called an Infinite Discontinuity.

4. Algebra of Continuous Functions

If $f$ and $g$ are two functions that are continuous at $x=a$, then:

• $f+g$, $f-g$, and $f \cdot g$ are continuous at $x=a$.
• $c \cdot f$ is continuous at $x=a$ for any constant $c$.
• $f/g$ is continuous at $x=a$, provided $g(a) \neq 0$.

5. Intermediate Value Theorem (IVT)

If a function $f(x)$ is continuous on a closed interval $[a, b]$ and $k$ is any number between $f(a)$ and $f(b)$ (where $f(a) \neq f(b)$), then there must exist at least one number $c$ in $(a, b)$ such that $f(c) = k$.

In simpler terms, a continuous function takes on all values between any two of its values.

[Image illustrating the Intermediate Value Theorem]

Solved Examples

Example 1: Checking for Continuity

Question: Discuss the continuity of the function $f(x) = \begin{cases} x+2, & x \leq 1 \\ x-2, & x > 1 \end{cases}$ at $x=1$.

Solution:

1. Value of the function: $f(1) = 1+2 = 3$. The function is defined at $x=1$.
2. Left-Hand Limit (LHL): $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x+2) = 1+2 = 3$$
3. Right-Hand Limit (RHL): $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x-2) = 1-2 = -1$$
4. Conclusion: Since LHL $\neq$ RHL ($3 \neq -1$), the limit $\lim_{x \to 1} f(x)$ does not exist. Therefore, the function is discontinuous at $x=1$. This is a jump discontinuity.

Example 2: Finding a Constant for Continuity

Question: Find the value of $k$ so that the function $f(x) = \begin{cases} kx^2, & x \leq 2 \\ 3, & x > 2 \end{cases}$ is continuous at $x=2$.

Solution: For the function to be continuous at $x=2$, we must have LHL = RHL = $f(2)$.

1. LHL: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (kx^2) = k(2^2) = 4k$$
2. RHL: $$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (3) = 3$$
3. Value of the function: $f(2) = k(2^2) = 4k$.
4. Equating: For continuity, LHL = RHL. $$4k = 3 \implies k = \frac{3}{4}$$

Example 3: Removable Discontinuity

Question: Show that the function $f(x) = \frac{x^2 - 4}{x - 2}$ for $x \neq 2$ has a removable discontinuity at $x=2$.

Solution:

1. Value of the function: $f(2)$ is not defined.
2. Limit at x=2: $$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x - 2} = \lim_{x \to 2} (x+2) = 4$$
3. Conclusion: The limit exists (LHL = RHL = 4), but the function is not defined at $x=2$. This is a removable discontinuity. We can remove it by defining a new function $g(x)$ as: $$g(x) = \begin{cases} \frac{x^2 - 4}{x - 2}, & x \neq 2 \\ 4, & x = 2 \end{cases}$$ This new function $g(x)$ is continuous at $x=2$.

Example 4: Continuity of Greatest Integer Function

Question: Discuss the continuity of $f(x) = [x]$ (the greatest integer function) at $x=3$.

Solution:

1. Value of the function: $f(3) = [3] = 3$.
2. LHL: As $x$ approaches 3 from the left, $x$ is slightly less than 3 (e.g., 2.999). $$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} [x] = 2$$
3. RHL: As $x$ approaches 3 from the right, $x$ is slightly greater than 3 (e.g., 3.001). $$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} [x] = 3$$
4. Conclusion: Since LHL $\neq$ RHL, the limit does not exist. The function $f(x)=[x]$ is discontinuous at every integer point. It exhibits a jump discontinuity at all integers.